Summation of Infinite Geometric Series, Part 2

By Chandrahas M. Halai

You might be wondering who must have been the first mathematician to calculate the sum of this infinite geometric series?

It was Virasena who had done so in his commentary Dhavala (816 CE) on the Digambara Jain text Shatkhandagama. He had calculated the sum of this infinite series to evaluate the volume of the frustum of a right circular cone.

This summation was proved by Nilakantha Somayaji (1444-1544 CE) (mathematician from Kerala in south India) in Aryabhatiyabhasya his commentary on Aryabhatiya. For details refer my earlier article:

https://chandrahasblogs.wordpress.com/2021/07/25/summation-of-infinite-geometric-series/

In this article I am going to derive the sum of the given infinite geometric series by geometric means.

Diagram 1

Step 1

Draw an equilateral triangle having an area 1 as shown in Diagram 1.

Diagram 2

Step 2

Join the midpoints of the sides of the equilateral triangle as shown in Diagram 2. This will divide the triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four triangles will be 1/4.

Diagram 3

Step 3

Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 3. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4^th of ¼.

Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be

Diagram 4

Step 4

Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 4. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4^th of .

Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be

Diagram 5

Step 5

Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 5. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4^th of .

Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be

Diagram 6

Step 6

Now join the midpoints of the sides of the top (white) triangle as shown in Diagram 6. This will divide the top triangle into four congruent (same size) equilateral triangles. Now the areas of each of these four new triangles will be 1/4^th of .

Now, the sum of the areas of each of the coloured triangles (blue, black and green) will be

Step 7

Repeat the above procedure ad infinitum.

Then, the sum of the areas of each of the coloured triangles (blue, black and green) will be

We can observe from diagram 6 that more the terms you add to the series the closer it gets to 1/3. Hence, we can say that the given series converges to 1/3.

Virahanka Sankhyanks in the Meru Prastaar

By Chandrahas M. Halai

This is in continuation of my earlier articles:

Meru Prastaar (मेरु प्रस्तारः)

 

Virahanka Numbers – Part I

 

https://www.indica.today/tag/virahanka-series/

What we popularly call as Fibonacci numbers are predated by Virahanka sankhyas.

In this article we derive a direct combinatorial formula to calculate the Virahanka sankhya.

 

Refer Table 1. This is left justified Meru Prastaar. If we add up the numbers in the shallow diagonals (represented by the lines in Table 1) we get Virahanka Sankhyanks.

The shallow diagonal that starts from row 0, gives us the 0^th Virahanka sankhya, which is 1. The shallow diagonal that starts from row 1, gives us the 1^st Virahanka sankhya, which is 1.

The shallow diagonal that starts from row 2, gives us the 2^nd Virahanka sankhya. This is = 1 + 1 = 2.

The shallow diagonal that starts from row 3, gives us the 3^rd Virahanka sankhya. This is = 1 + 2 = 3.

The shallow diagonal that starts from row 4, gives us the 4^th Virahanka sankhya. This is = 1 + 3 + 1 = 5.

The shallow diagonal that starts from row 5, gives us the 5^th Virahanka sankhya. This is = 1 + 4 + 3 = 8.

The shallow diagonal that starts from row 6, gives us the 6^th Virahanka sankhya. This is = 1 + 5 + 6 +1 = 13.

And so on.

Refer Table 2. This gives the combinatorial formula for the binomial coefficients. Using Table 2, we can calculate the Virahanka sankhyas using the combinatorial formulas. Thus we have:

Do we discern a pattern here? Can we have a formula to calculate the Virahanka sankhyank for a quarter having n matras?

Let us consider a quarter having 5 matras. The metrical forms in a 5 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3 and 5. To put it another way, the metrical forms in a 5 matra prastaar will have 0, 1 and 2 Gurus. The metrical forms in a 5 matra prastaar will have at least 1 Laghu syllable.

The total metrical forms in a 5 matra prastaar are

If there is 1 Guru in the metrical forms then there will be 5 – 1 = 4 syllables in the quarter. If there are 2 Gurus in the metrical forms then there will be 5 – 2 = 3 syllables in the quarter.

In general, for a n matra prastaar, if there are r Gurus in the metrical forms then there will be n – r syllables in the quarter.

Let us now consider a quarter having 6 matras. The metrical forms in a 6 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4 and 6. To put it another way, the metrical forms in a 6 matra prastaar will have 0, 1, 2 and 3 Gurus.

The total metrical forms in a 6 matra prastaar are

Let us now consider a quarter having 7 matras. The metrical forms in a 7 matra prastaar will only have odd numbers of Laghus, i.e. 1, 3, 5 and 7. To put it another way, the metrical forms in a 7 matra prastaar will have 0, 1, 2 and 3 Gurus. The metrical forms in a 7 matra prastaar will have at least 1 Laghu syllable.

The total metrical forms in a 7 matra prastaar are

Let us now consider a quarter having 8 matras. The metrical forms in a 8 matra prastaar will only have even numbers of Laghus, i.e. 0, 2, 4, 6 and 8. To put it another way, the metrical forms in a 8 matra prastaar will have 0, 1, 2, 3 and 4 Gurus.

The total metrical forms in a 8 matra prastaar are

From above, we observe that:

In general in a n matra vrutta prastaar, if n is even the metrical forms will have only even number of Laghus. And if n is odd the metrical forms will have only odd number of Laghus.

In general in a n matra vrutta prastaar, if n is even the metrical forms can have maximum n/2 number of Gurus. And if n is odd the metrical forms can have maximum (n – 1)/2 number of Gurus.

Therefore, the total metrical forms in a n matra prastaar are:

1. If n is even
   

   

   
   

   The number of terms that are added in this formula will be n/2 + 1.
   

    

2. If n is odd
   

   

The number of terms that are added in this formula will be (n + 1)/2.

Thus we now have a direct combinatorial formula for calculating the Virahanka sankhyas.